# Determine the magnitude of the reactions

Determine the magnitude of the reactions on the beam at A and B. Neglect the thickness of the beam.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

### Solution:

Let us first draw a free body diagram showing all the forces like so:

We can now write a moment equation at point A to determine force $B_y$:

$\circlearrowright^+ M_A=0;$

$600(4)+400\text{cos}\,(15^0)(12)-B_y(12)=0$

$B_y=586.37$ N

Now, we can write an equilibrium equation for the x-axis forces.

$\rightarrow^+\Sigma F_x=0;$

$A_x-400\text{sin}\,(15^0)=0$

$A_x=103.53$ N

Lastly, we can write an equilibrium equation for the y-axis forces.

$\uparrow^+ \Sigma F_y=0;$

$A_y+B_y-600-400\text{cos}\,(15^0)=0$

###### (Remember we already found $B_y=586.37$ N)

$A_y=400$ N

Now that we know $A_x$ and $A_y$, we can find the magnitude. Remember, magnitude is:

$F_A=\sqrt{(A_x)^2+(A_y)^2}$

$F_A=\sqrt{(103.53)^2+(400)^2}$

$F_A=413.18$ N

$F_A=413.18$ N
$F_B=B_y=586.37$ N